A、Rescue The Princess
已知一个等边三角形的两个顶点A、B，求第三个顶点C，A、B、C成逆时针方向。
常规的解题思路就是用已知的两个点列出x，y方程，但这样求出方程的解的表达式比较复杂。更为简单的方法就是以A的坐标加A、C直线的角度求解，使用atan2函数求出A、B直线的角度再加上60就是A、C的角度，使用A、B求出等边三角形的边长。
1 #include <cstdio> 2 #include <cmath> 3 #include <iostream> 4 5 using namespace std; 6 7 const double pi = acos(-1.0); 8 double xx1,yy1,xx2,yy2,xx3,yy3,jiao,leng; 9 10 int main(int argc, char const *argv[]) 11 { 12 freopen("SD_3230_Rescue_The_Princess.in","r",stdin); 13 int amount; 14 scanf("%d",&amount); 15 while(amount--){ 16 scanf("%lf %lf %lf %lf",&xx1,&yy1,&xx2,&yy2); 17 jiao = atan2(yy2-yy1,xx2-xx1); 18 leng = sqrt(pow(xx1-xx2,2)+pow(yy1-yy2,2)); 19 xx3 = xx1 + leng*cos(jiao+pi/3.0); 20 yy3 = yy1 + leng*sin(jiao+pi/3.0); 21 printf("(%.2lf,%.2lf) ", xx3,yy3); 22 } 23 return 0; 24 }
B、Thrall’s Dream
给出N个点N (0 <= N < 2001)，M条边M (0 <= M < 10001)构成的图，判断是否任意两点可达，可达输出“Kalimdor is just ahead“，否则输出”The Burning Shadow consume us all“。
首先拿到这题第一个出现脑海的想法就是一次对每个点进行dfs，并记录已经确定的结果。但是想一想肯定会TE，所以就此打住这个想法。采用targan算法求出所有的连通分量，再判断这些连通分量能否构成一条链（不能出现两个或两个以上出度为0的连通分量），能则true，否则false，关键在于使用targan算法求出所有的联通分量。targan算法：基于dfs，使用low记录该点所属连通分量，dfn记录遍历到此点的时间，如果遍历的点已经遍历过并在栈中说明出现了环，要更新的相应的low值到最小，并把小的low值回传给father，如果出现low==dfn则说明出现了一个联通分量，并且这个连通分量的所有元素一定都在栈中并且都相邻，然后将这些元素全部出栈并统计此连通分量的出度，最后如果有大于等于两个连通分量的出度为0则一定有连点不可达。
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <stack> 5 6 #ifndef SYMBOL 7 #define MAXN 2001 8 #endif 9 10 using namespace std; 11 /* 12 dfs变形的targan算法，求强连通图的强联通分量 13 */ 14 15 struct node 16 { 17 int size; 18 int indegree; 19 int asked; 20 int instack; 21 int father; 22 }; 23 24 int childs[MAXN][MAXN],low[MAXN],dfn[MAXN]; 25 int nodeAmount,edgeAmount,times,mins,outzero; 26 stack<int> stacks; 27 struct node nodes[MAXN]; 28 29 void init(){ 30 for (int i = 1;i <= MAXN;i ++){ 31 nodes[i].size = 0; 32 nodes[i].asked = 0; 33 nodes[i].instack = 0; 34 nodes[i].indegree = 0; 35 } 36 times = 0; 37 outzero = 0; 38 mins = 0; 39 memset(childs,0,sizeof(childs)); 40 memset(low,0,sizeof(low)); 41 memset(dfn,0,sizeof(dfn)); 42 memset(nodes,0,sizeof(nodes)); 43 while(!stacks.empty()){ 44 stacks.pop(); 45 } 46 } 47 48 void targan(int n){ 49 times ++; 50 dfn[n] = times; 51 low[n] = times; 52 stacks.push(n); 53 nodes[n].asked = 1; 54 nodes[n].instack = 1; 55 int child; 56 mins = dfn[n]; 57 for (int i = 0;i < nodes[n].size;i ++){ 58 child = childs[n][i]; 59 nodes[child].father = n; 60 if (nodes[child].instack){ 61 mins = std::min(mins ,dfn[child]); 62 mins = std::min(mins ,low[child]); 63 low[n] = mins; 64 } 65 if (!nodes[child].asked){ 66 targan(child); 67 } 68 } 69 int father = nodes[n].father; 70 mins = low[father]; 71 mins = std::min(mins ,dfn[n]); 72 mins = std::min(mins ,low[n]); 73 low[father] = mins; 74 if (dfn[n] == low[n]){ 75 int nod; 76 bool isout = false; 77 while(1){ 78 nod = stacks.top(); 79 stacks.pop(); 80 nodes[nod].instack = 0; 81 int size = nodes[nod].size; 82 int chil; 83 for (int k = 0;k < size;k ++){ 84 chil = childs[nod][k]; 85 if (low[chil] != low[nod]){ 86 isout = true; 87 } 88 } 89 if (nod == n) break; 90 } 91 if (!isout){ 92 outzero ++; 93 } 94 } 95 } 96 97 bool judge(){ 98 if (outzero >= 2){ 99 return false; 100 } 101 return true; 102 } 103 104 int main(){ 105 freopen("SD_3231_Thralls_Dream.in","r",stdin); 106 int amount; 107 scanf("%d",&amount); 108 for (int i = 1;i <= amount;i ++){ 109 scanf("%d %d",&nodeAmount,&edgeAmount); 110 init(); 111 int from,to; 112 for (int j = 0;j < edgeAmount;j ++){ 113 scanf("%d %d",&from,&to); 114 int size = nodes[from].size; 115 childs[from][size++] = to; 116 nodes[from].size = size; 117 nodes[to].indegree ++; 118 } 119 for (int k = 1;k <= nodeAmount;k ++){ 120 if (!nodes[k].asked){ 121 targan(k); 122 } 123 } 124 if (judge()) { 125 printf("Case %d: Kalimdor is just ahead ", i); 126 } else { 127 printf("Case %d: The Burning Shadow consume us all ",i); 128 } 129 } 130 return 0; 131 }
f(x) = K, x = 1，f(x) = (a*f(x-1) + b)%m , x > 1，求( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P. 1 <= n <= 10^6，0 <= A, K, a, b <= 10^9，1 <= m, P <= 10^9
10^9无论从空间还是时间上都是不可行的，所以简单的计算绝对不行，A^(10^9)我们可以预处理一下，求A^(ki+j)直接将10^9的复杂度降了一个平方。只需求出A^k即可
1 #include <cstdio> 2 #include <cmath> 3 #include <iostream> 4 5 #ifndef SYMBOL 6 #define MAXK 100005 7 #endif 8 9 typedef long long ll; 10 11 using namespace std; 12 /* 13 合理的运用%的特性 14 预处理(A^1~n)%P次方 15 */ 16 17 ll n, A, K, a, b, m, P; 18 ll pre1[MAXK+1],pre2[MAXK+1]; 19 20 void init(){ 21 pre1[0] = pre2[0] = 1LL; 22 for (int i = 1;i <= MAXK;i ++){ 23 pre1[i] = A*pre1[i-1]%P; 24 } 25 for (int i = 1;i <= MAXK;i ++){ 26 pre2[i] = pre2[i-1]*pre1[MAXK]%P; 27 } 28 } 29 30 ll gao(){ 31 ll res = 0,t = K; 32 for (int i = 1;i <= n;i ++){ 33 res = (res + pre2[t/MAXK]*pre1[t%MAXK])%P; 34 t = (a*t + b)%m; 35 } 36 return res; 37 } 38 39 int main(int argc, char const *argv[]) 40 { 41 freopen("SD_3232_AX_mod_P.in","r",stdin); 42 int test; 43 scanf("%d",&test); 44 for (int i = 1;i <= test;i ++){ 45 scanf("%lld %lld %lld %lld %lld %lld %lld",&n,&A,&K,&a,&b,&m,&P); 46 init(); 47 printf("Case #%d: %lld ", i,gao()); 48 } 49 return 0; 50 }
D、Rubik’s cube
求将一个两色四方格旋转致每面都是同一颜色的最小步数。
直接模拟旋转魔方进行bfs，魔方可以前后，左右，上下共有12种旋转方案，一面的向前旋转和另一面的向后旋转是同一效果，除以2剩6种旋转可能，向后旋转一下相当于向前旋转3下，除以2剩3种可能的旋转方案，再依次对三种旋转进行bfs
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <queue> 5 #include <map> 6 7 #ifndef SYMBOL 8 #define CUBE 24 9 #define ONLINE_JUDGE 1 10 #endif 11 12 using namespace std; 13 14 struct node 15 { 16 int cu[CUBE]; 17 int step; 18 }; 19 20 node start; 21 queue<node> que; 22 map<int,int> myhash; 23 int rode[6][8] = { 24 {4,6,17,16,15,13,9,8}, 25 {1,3,2,0}, 26 {1,3,17,19,22,20,10,8}, 27 {4,6,7,5}, 28 {0,1,4,5,20,21,12,13}, 29 {9,8,10,11} 30 }; 31 32 void turn(node & n,int face){ 33 face<<=1; 34 int t = n.cu[rode[face][0]]; 35 for (int i = 0;i < 7;i ++){ 36 n.cu[rode[face][i]] = n.cu[rode[face][i+1]]; 37 } 38 n.cu[rode[face][7]] = t; 39 t = n.cu[rode[face][0]]; 40 for (int i = 0;i < 7;i ++){ 41 n.cu[rode[face][i]] = n.cu[rode[face][i+1]]; 42 } 43 n.cu[rode[face][7]] = t; 44 face ++; 45 t = n.cu[rode[face][0]]; 46 for (int i = 0;i < 3;i ++){ 47 n.cu[rode[face][i]] = n.cu[rode[face][i+1]]; 48 } 49 n.cu[rode[face][3]] = t; 50 } 51 52 int myhashCode(const node n){ 53 int res = 0; 54 for (int i = 0;i < 24;i ++){ 55 res = n.cu[i] + (res<<1); 56 } 57 return res; 58 } 59 60 bool isOK(const node n){ 61 for (int i = 0;i < 24;i += 4){ 62 for (int j = 1;j <= 3;j ++){ 63 if (n.cu[i] != n.cu[i+j]){ 64 return false; 65 } 66 } 67 } 68 return true; 69 } 70 71 int bfs(){ 72 int red = 0; 73 for (int i = 0;i < 24;i ++){ 74 if (start.cu[i] == 0){ 75 red ++; 76 } 77 } 78 if (red%4 != 0){ 79 return -1; 80 } 81 if (isOK(start)){ 82 return start.step; 83 } 84 myhash.clear(); 85 while(!que.empty()) que.pop(); 86 myhash[myhashCode(start)] = 1; 87 que.push(start); 88 node s,t; 89 while(!que.empty()){ 90 s = que.front(); 91 que.pop(); 92 for (int i = 0;i < 3;i ++){ 93 t = s; 94 t.step ++; 95 turn(t,i); 96 if (myhash.find(myhashCode(t)) == myhash.end()){ 97 if (isOK(t)){ 98 return t.step; 99 } 100 myhash[myhashCode(t)] = 1; 101 que.push(t); 102 } 103 turn(t,i); 104 turn(t,i); 105 if (myhash.find(myhashCode(t)) == myhash.end()){ 106 if (isOK(t)){ 107 return t.step; 108 } 109 myhash[myhashCode(t)] = 1; 110 que.push(t); 111 } 112 } 113 } 114 return -1; 115 } 116 117 int main(int argc, char const *argv[]) 118 { 119 #ifndef ONLINE_JUDGE 120 freopen("SD_3232_Rubiks_cube.in","r",stdin); 121 #endif 122 int test; 123 scanf("%d",&test); 124 while(test--){ 125 for (int i = 0;i < 24;i ++){ 126 scanf("%d",&start.cu[i]); 127 } 128 start.step = 0; 129 int res = bfs(); 130 if (res == -1) printf("IMPOSSIBLE! "); 131 else printf("%d ", res); 132 } 133 return 0; 134 }
E、Mountain Subsequences
在一个数字序列中求山脉序列的个数 a1 < ...< ai < ai+1 < Amax > aj > aj+1 > ... > an
找出比当前数字前面并比当前数字小的数进行dp
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define MAX 100000 5 #define MOD 2012 6 using namespace std; 7 //cnt记录单个字符的个数，dp记录以此字母结束的个数 8 int leng,up[MAX],down[MAX],dp[26],cnt[26],t; 9 char str[MAX]; 10 int main() { 11 freopen("SD_3234_Mountain_Subsequences.in","r",stdin); 12 while(~scanf("%d",&leng)){ 13 scanf("%s",str); 14 memset(up,0,sizeof(up)); 15 memset(down,0,sizeof(down)); 16 memset(dp,0,sizeof(dp)); 17 memset(cnt,0,sizeof(cnt)); 18 for (int i = 0;i < leng;i ++){ 19 t = str[i] - 'a'; 20 for (int j = 0;j < t;j ++){ 21 up[i] += dp[j] + cnt[j]; 22 up[i] %= MOD; 23 } 24 cnt[t] ++; 25 cnt[t] %= MOD; 26 dp[t] += up[i]; 27 dp[t] %= MOD; 28 } 29 memset(dp,0,sizeof(dp)); 30 memset(cnt,0,sizeof(cnt)); 31 for (int i = leng-1;i >= 0;i --){ 32 t = str[i] - 'a'; 33 for (int j = 0;j < t;j ++){ 34 down[i] += dp[j] + cnt[j]; 35 down[i] %= MOD; 36 } 37 cnt[t] ++; 38 cnt[t] %= MOD; 39 dp[t] += down[i]; 40 dp[t] %= MOD; 41 } 42 int sum = 0; 43 for (int i = 1;i < leng-1;i ++){ 44 sum += up[i]*down[i]; 45 sum %= MOD; 46 } 47 printf("%d ", sum); 48 } 49 return 0; 50 }
F、Alice and Bob
(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)，求x^P的系数
1 <= P <= 2^0+2^1+2^2+.....+2^(n-1)，二次项有个性质2^0+2^1+......+2^(n-1)<2^n，所以可以推断出P只能由唯一的二次项系数构成。将P的二进制求出来，将其中值为1的项的系数相乘即为结果。
1 #include <cstdio> 2 #include <iostream> 3 #include <cmath> 4 #include <cstring> 5 6 #ifndef SYMBOL 7 #define MAX 55 8 #endif 9 10 using namespace std; 11 /* 12 定理：一个数只能由2的多项式中唯一的序列构成2^0 2^1 2^2 2^3 2^4 ,7只能由2^0 2^1 2^2构成 13 */ 14 15 typedef long long ll; 16 17 int n,dex; 18 int co[MAX],binary[MAX]; 19 20 void getBinary(ll number){ 21 memset(binary,0,sizeof(binary)); 22 dex = 0; 23 while(number){ 24 binary[dex++] = number%2LL; 25 number /= 2LL; 26 } 27 } 28 29 int calcu(ll P){ 30 if (P == 0LL){ 31 return 1;//一定要考虑极限问题 32 } 33 ll max = 0; 34 for (int i = 0;i < n;i ++){ 35 max += pow(2LL,i); 36 } 37 if (P > max){ 38 return 0;//一定要考虑极限问题 39 } 40 getBinary(P); 41 int sum = 1; 42 for (int i = 0;i < dex;i ++){ 43 if (binary[i]){ 44 sum *= co[i]; 45 sum %= 2012; 46 } 47 } 48 return sum; 49 } 50 51 int main(int argc, char const *argv[]) 52 { 53 // freopen("SD_3235_Alice_and_Bob.in","r",stdin); 54 // freopen("SD_3235_Alice_and_Bob.out","w",stdout); 55 int test,ques; 56 ll P; 57 scanf("%d",&test); 58 while(test--){ 59 scanf("%d",&n); 60 for (int i = 0;i < n;i ++){ 61 scanf("%d",&co[i]); 62 } 63 scanf("%d",&ques); 64 while(ques--){ 65 scanf("%lld",&P); 66 printf("%d ",calcu(P)); 67 } 68 } 69 return 0; 70 }
G、A-Number and B-Number
可以被7整除或包含7的为A数字，A数字中下标为A数字的除外其余为B数字，求第n个B数字。
对位数进行dp
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 5 #ifndef SYMBOL 6 #define MAX 25 7 #endif 8 9 /* 10 位数动态规划,设定最大25位,dp的方式有很多 11 */ 12 using namespace std; 13 14 typedef unsigned long long ull; 15 16 ull number; 17 ull dp[MAX][7][2]; 18 int bits[MAX]; 19 20 ull dfs(int d,int mod7,int have7,bool limit){ 21 if (!d){ 22 return (!mod7 || have7); 23 } 24 if (!limit && ~dp[d][mod7][have7]){ 25 return dp[d][mod7][have7]; 26 } 27 int mn = 9; 28 if (limit){ 29 mn = bits[d]; 30 } 31 ull sum = 0ULL; 32 for (int i = 0;i <= mn;i ++){ 33 int tmod7 = (10*mod7+i)%7; 34 int thave7 = have7 || (i == 7); 35 sum += dfs(d-1,tmod7,thave7,(i == mn) && limit); 36 } 37 if (!limit){ 38 dp[d][mod7][have7] = sum; 39 } 40 return sum; 41 } 42 43 ull AAmount(ull n){ 44 int dex = 0; 45 while(n){ 46 bits[++dex] = n%10; 47 n/=10; 48 } 49 return dfs(dex,0,0,true)-1; 50 } 51 52 ull twoSplite(){ 53 ull l = 7ULL,r = (1ULL<<63) - 1,m; 54 while(l <= r){ 55 m = (l+r)>>1; 56 ull amount = AAmount(m); 57 amount = amount - AAmount(amount); 58 if (amount < number){ 59 l = m+1; 60 }else { 61 r = m-1; 62 } 63 //注意这里==不能返回 64 } 65 return l; 66 } 67 68 int main(int argc, char const *argv[]) 69 { 70 freopen("SD_3236_A-Number_and_B-Number.in","r",stdin); 71 memset(dp,-1,sizeof(dp)); 72 while(~scanf("%llu",&number)){ 73 printf("%llu ", twoSplite()); 74 } 75 return 0; 76 }
H、Boring Counting
求一定范围内一定大小范围内的数的个数。
采用线段树数据结构
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 5 #ifndef SYMBOL 6 #define MAX 50005 7 #endif 8 9 /*无形的线段树*/ 10 using namespace std; 11 12 struct num 13 { 14 int value; 15 int dex; 16 bool operator <(const num n) const {return value < n.value;}; 17 }nums[MAX]; 18 19 struct query 20 { 21 int l; 22 int r; 23 int a; 24 int b; 25 int dex; 26 }querys[MAX]; 27 28 bool comp1(const query q1,const query q2) { 29 return q1.a < q2.a; 30 } 31 32 bool comp2(const query q1,const query q2) { 33 return q1.b < q2.b; 34 } 35 36 int nn,qn; 37 int minsum[MAX<<2],ans[MAX][2]; 38 /*ans[MAX][0]记录比a小数的个数，ans[MAX][1]记录比b小数的个数，答案为ans[MAX][1]-ans[MAX][0]*/ 39 40 /*创建线段树,额外添加的问题值为在这个范围内的个数*/ 41 void build(int l,int r,int dex){ 42 if (l == r){ 43 minsum[dex] = 0; 44 return ; 45 } 46 int m = (l+r)>>1; 47 build(l,m,dex<<1); 48 build(m+1,r,dex<<1|1); 49 minsum[dex] = minsum[dex<<1] + minsum[dex<<1|1]; 50 } 51 52 void update(int d,int l,int r,int dex){ 53 if (l == r){ 54 minsum[dex] ++; 55 return ; 56 } 57 int m = (l+r)>>1; 58 if (d <= m){ 59 update(d,l,m,dex<<1); 60 }else{ 61 update(d,m+1,r,dex<<1|1); 62 } 63 minsum[dex] = minsum[dex<<1] + minsum[dex<<1|1]; 64 } 65 66 int queryy(int ql,int qr,int l,int r,int dex){ 67 if (ql <= l && qr >= r){ 68 return minsum[dex]; 69 } 70 int m = (l+r)>>1; 71 int res = 0; 72 if (ql <= m){ 73 res += queryy(ql,qr,l,m,dex<<1); 74 } 75 if (qr > m){ 76 res += queryy(ql,qr,m+1,r,dex<<1|1); 77 } 78 return res; 79 } 80 81 void calculate(){ 82 build(1,nn,1); 83 sort(nums+1,nums+nn+1); 84 sort(querys+1,querys+qn+1,comp1); 85 int dd = 1; 86 for (int j = 1;j <= qn;j ++){ 87 while(dd <= nn && nums[dd].value < querys[j].a){//注意这里等于不能包括在内 88 update(nums[dd].dex,1,nn,1); 89 dd++; 90 } 91 ans[querys[j].dex][0] = queryy(querys[j].l,querys[j].r,1,nn,1); 92 } 93 build(1,nn,1); 94 sort(querys+1,querys+qn+1,comp2); 95 dd = 1; 96 for (int j = 1;j <= qn;j ++){ 97 while(dd <= nn && nums[dd].value <= querys[j].b){ 98 update(nums[dd].dex,1,nn,1); 99 dd++; 100 } 101 ans[querys[j].dex][1] = queryy(querys[j].l,querys[j].r,1,nn,1); 102 } 103 } 104 105 void print(int cases){ 106 printf("Case #%d: ",cases); 107 for (int i = 1;i <= qn;i ++){ 108 printf("%d ",ans[i][1]-ans[i][0]); 109 } 110 } 111 112 int main(int argc, char const *argv[]) 113 { 114 freopen("SD_3237_Boring_Counting.in","r",stdin); 115 int test; 116 scanf("%d",&test); 117 for (int i = 1;i <= test;i ++){ 118 scanf("%d %d",&nn,&qn); 119 for (int j = 1;j <= nn;j ++){ 120 scanf("%d",&nums[j].value); 121 nums[j].dex = j; 122 } 123 for (int j = 1;j <= qn;j ++){ 124 scanf("%d %d %d %d",&querys[j].l,&querys[j].r,&querys[j].a,&querys[j].b); 125 querys[j].dex = j; 126 } 127 calculate(); 128 print(i); 129 } 130 return 0; 131 }
I、The number of steps
递推
1 #include <cstdio> 2 #include <iostream> 3 4 #define maxn 55 5 6 using namespace std; 7 8 double dp[maxn][maxn]; 9 10 int main() 11 { 12 // freopen("SD_3238_The_number_of_steps.in","r",stdin); 13 int n; 14 while(scanf("%d",&n),n) 15 { 16 double a,b,c,d,e; 17 scanf("%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e); 18 dp[n][0]=0; 19 for(int i=1;i<=n;++i) 20 dp[n][i]=dp[n][i-1]+1.0; 21 for(int i=n-1;i>0;--i) 22 { 23 dp[i][0]=a*dp[i+1][0]+b*dp[i+1][1]+1.0; 24 for(int j=1;j<n;++j) 25 dp[i][j]=c*dp[i+1][j]+d*dp[i+1][j+1]+e*dp[i][j-1]+1.0; 26 } 27 printf("%.2lf ",dp[1][0]); 28 } 29 return 0; 30 }
J、Contest Print Server
模拟
1 #include <cstdio> 2 #include <iostream> 3 4 #ifndef SYMBOL 5 #define MAX 100 6 #endif 7 8 using namespace std; 9 10 struct task 11 { 12 char name[30]; 13 int page; 14 }tasks[MAX]; 15 16 int n,s,x,y,mod; 17 18 19 void print(){ 20 int task = 0, sum = 0; 21 while(task < n){ 22 if (sum == s){ 23 printf("%d pages for %s ",0,tasks[task].name); 24 s=(s*x+y)%mod; 25 sum = 0; 26 }else if (sum + tasks[task].page <= s){ 27 printf("%d pages for %s ",tasks[task].page,tasks[task].name); 28 sum += tasks[task].page; 29 task ++; 30 }else{ 31 printf("%d pages for %s ",s - sum,tasks[task].name); 32 sum = 0; 33 s=(s*x+y)%mod; 34 } 35 } 36 } 37 38 int main(int argc, char const *argv[]) 39 { 40 // freopen("SD_3239_Contest Print_Server.in","r",stdin); 41 // freopen("SD_3239_Contest Print_Server.out","w",stdout); 42 int test; 43 scanf("%d",&test); 44 while(test--){ 45 scanf("%d %d %d %d %d",&n,&s,&x,&y,&mod); 46 for (int i = 0;i < n;i ++){ 47 scanf("%s request %d pages",tasks[i].name,&tasks[i].page); 48 } 49 print(); 50 printf(" "); 51 } 52 return 0; 53 }